Weight,
Mass, and the
Dreaded
Elevator
Problem
Mini-lab: Weight vs. Mass

Determine the mathematical relationship
between an object’s weight and its mass.
– Materials:
 Electronic Scales
 Triple Beam Balance
 Multiple objects of different mass
Create a data table BEFORE starting the lab
 Follow the instructions at the lab station.
 Create a graph of force vs. mass
 Compare the slope of your graph to g

Weight: True or False?
In your journal, re-write the statements
below, state whether you think it is true or
false, and provide 1-2 sentences of your
reasoning.
– The mass of an object depends on its
location
– The weight of an object depends on its
location
– Mass and weight are the same, but with
different units.
Weight vs. Mass

Weight: The force that gravity exerts on
an object with mass (m).
– This force is what causes falling bodies to
accelerate at 9.80 m/s2.
– Weight is ALWAYS directed toward the center
of the earth (down)
Fg  m  g
– Remember, g = 9.80 m/s2
– Units = Newtons
Apparent Weight
Apparent weight is the weight
something appears to have as a result of
an acceleration.
 For example, if you were standing on a
scale in an elevator, your apparent weight
is the weight the scale would read.


So now for some conceptual practice…
Apparent Weight Practice…

Suppose you have a jet-powered flying platform
that can move straight up and down. For each
of the following cases, is you apparent weight
equal to, greater than, or less then your true
weight? Explain.
–
–
–
–
–
–
You
You
You
You
You
You
are
are
are
are
are
are
ascending and speeding up
descending and speeding up
ascending at a constant speed
ascending and slowing down
descending and slowing down
descending at a constant speed
Heavier
Lighter
Same
Constant Vertical Velocity
Example: A leaf falling at terminal velocity
 Up is the positive (+) direction
• 𝑎 = 0 so … 𝐹𝑁𝐸𝑇 = 0
• 𝐹𝑁𝐸𝑇 = 0 = 𝐹𝑢𝑝 − 𝐹𝑔
+𝐹𝑔
𝐹𝑢𝑝 = 𝐹𝑔
+ 𝐹𝑔
(use the definition of weight)
𝐹𝑢𝑝 = 𝑚𝑔
F up = Fair
Fg
Calculating Apparent Weight
Apparent weight can easily be calculated
using the concept of net force.
 For example, if you are standing on a
scale when you are at rest, what forces
are acting on you?

– The force of gravity (your weight) and the force of
the scale pushing back up (the normal force)

What is the net force in this situation?
– 0 N … you’re in static equilibrium

Draw a Free-body diagram for this
situation:
Fscale = apparent weight
Fg = m g

Write out the vector equation:
F net  F scale  F g
F net  F scale  F g

Since this situation is in equilibrium,
Fnet  0

F scale  F g
Therefore,
which means the scale is reading the
“True weight”
If the person standing on the scale has a
mass of 65.0 kg, what is his weight?
F g  m  g  ( 65 . 0 kg )( 9 . 80 m
s
2
)  637 N
Accelerating Upwards
Example: A crate being lifted by a rope
 Up is the positive (+) direction
• 𝑎 is ↑ so … 𝐹𝑁𝐸𝑇 is ↑
• 𝐹𝑁𝐸𝑇 = 𝐹𝑢𝑝 − 𝐹𝑔
F up = FT
+𝐹𝑔
+ 𝐹𝑔
𝐹𝑁𝐸𝑇 + 𝐹𝑔 = 𝐹𝑢𝑝
Fg
𝑚𝑎 + 𝑚𝑔 = 𝐹𝑢𝑝 (using 2
nd
Law and the definition of weight)
𝐹𝑢𝑝 = 𝑚 𝑎 + 𝑔
An Accelerating elevator…

If the elevator is accelerating upwards or
downwards, then our problem becomes
slightly longer…

For example, let’s say the elevator is
accelerating upwards at a rate of 2.00 m/s2.
What is now different from our first
example?

Draw a free body diagram, including a
vector off to the side indicating the
direction of the net force:
Fscale = apparent weight
Fnet
Fg = m g

Then write the vector equation:
F net  F scale  F g
F net  F scale  F g

Since this situation is NOT in equilibrium,
the following is ALSO TRUE:
Fnet  ma

Using substitution, we can determine the
size of the apparent weight (the reading
on the scale):
ma  F scale  mg
F scale  ma  mg  m ( a  g )
F scale  ( 65 . 0 kg )( 2 . 00 m
s
2
 9 . 80 m
s
2
)  767 N
Accelerating Downwards
Example: A sky diver in free fall
 Up is the positive (+) direction
• 𝑎 is ↓ so … 𝐹𝑁𝐸𝑇 is ↓
• −𝐹𝑁𝐸𝑇 = 𝐹𝑢𝑝 − 𝐹𝑔
F up = Fair
Fg
+𝐹𝑔
+ 𝐹𝑔
−𝐹𝑁𝐸𝑇 + 𝐹𝑔 = 𝐹𝑢𝑝
−𝑚𝑎 + 𝑚𝑔 = 𝐹𝑢𝑝 (using 2
nd
Law and the definition of weight)
𝐹𝑢𝑝 = 𝑚 𝑔 − 𝑎
Another Accelerating elevator…

Now let’s say the elevator is accelerating
downwards at a rate of 2.00 m/s2. Draw
the free-body diagram for this situation:
Fscale = apparent weight
Fnet
Fg = m g

Again, we can write the vector equations…
HOWEVER: the net force is now DOWN, so it (and
the acceleration) is therefore a negative value…
 F net  F scale  mg
 F net  m (  a )

Using substitution, we can determine the size of
the apparent weight (the reading on the scale):
m (  a )  F scale  mg
F scale  m (  a )  mg  m (  a  g )  m ( g  a )
F scale  ( 65 . 0 kg )( 9 . 80 m
s
2
 2 . 00 m
s
2
)  507 N
Now You Try!

Determine the apparent weight of a 67 kg man
standing in an elevator when the elevator is:
– At rest
F g  m  g  ( 67 kg )( 9 . 80 m
s
2
)  660 N
– Ascending and speeding up at a rate of 1.5 m/s2
F scale  ma  mg  m ( a  g )
F scale  ( 67 kg )(1 . 5 m
s
2
 9 . 80 m
s
2
)  760 N
– Ascending and slowing down at a rate of -1.2 m/s2
F scale  m (  a )  mg  m (  a  g )  m ( g  a )
F scale  ( 67 kg )( 9 . 80 m
s
2
 1 .2 m
s
2
)  580 N